How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$?

Question

I was given the series: $$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$ Making some observations I realized that the $ a_{n} $ term would be the following: $$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$ What I wanted to do is to find the result of the series, so the answer would be: $$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$ I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.

Answer 1

HINT:

$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$

Separate odd and even terms.

$\frac{1}{3} +\frac{1}{3^{3}} + \frac{1}{3^{5}} + ...$

$ \frac{2}{3^{2} } + \frac{2}{3^{4}} +\frac{2}{3^{6}} ...$

Answer 2

It is acceptable to split one convergent series in two as long as both are convergent.

$$ 0 < \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ... < \frac{2}{3} + \frac{2}{3^{2} } + \frac{2}{3^{3}} + \frac{2}{3^{4}} + \frac{2}{3^{5}} + ...$$

and since

$$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$ is monotonically increasing as you are adding terms, and it is bounded by another convergent series (above), it is convergent itself.

Now:

$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...=\sum_{n=1}^{\infty} \frac{1}{3^{2n-1}}+\sum_{n=1}^{ \infty }\frac{2}{3^{2n} }=3\sum_{n=1}^{\infty} \frac{1}{3^{2n}}+2\sum_{n=1}^{ \infty }\frac{1}{3^{2n} }=5\sum_{n=1}^{ \infty }\frac{1}{9^{n} }=\frac{5}{8}$$

Answer 3

Your sum can be put as

$$(1+\frac 23)\Bigl(\frac 13+\frac{1}{3^3}+\frac{1}{3^5}...\Bigr)=$$ $$\frac 53\frac 13\Bigl(1+\frac 19+\frac{1}{9^2}+...\Bigr)=$$ $$\frac 53\frac 13 \frac{1}{1-\frac 19}=\frac 58$$

Answer 4

There is a much simpler way to solve this. You know how to compute the geometric series $$\frac{1}{3} + \frac{1}{3^2}+ \frac{1}{3^3}+ \frac{1}{3^4}+ \cdots = \frac{1}{2}$$ you also know how to compute the geometric series $$\frac{1}{9} + \frac{1}{9^2}+ \frac{1}{9^3}+ \frac{1}{9^4}+ \cdots= \frac{1}{8}$$ But this can be rewritten in terms of $3^2$, so that you get $$\frac{1}{3^2} + \frac{1}{3^4}+ \frac{1}{3^6}+ \frac{1}{3^8}+ \cdots$$ Now the answer to your question is simply $\frac{1}{2} +\frac{1}{8}$.

Answer 5

The series can also be written as $$ \begin{align} \overbrace{\ \sum_{k=1}^\infty\frac1{3^k}\ }^{\substack{\text{each power}\\\text{of $3$}}}+\overbrace{\ \sum_{k=1}^\infty\frac1{9^k}\ }^{\substack{\text{even powers}\\\text{of $3$}}} &=\frac{\frac13}{1-\frac13}+\frac{\frac19}{1-\frac19}\\ &=\frac12+\frac18\\[6pt] &=\frac58 \end{align} $$

Answer 6

In base $3$, we can say

$$N = \frac {1}{10} + \frac{2}{100} + \frac{1}{1000} + \frac{2}{10000} + \cdots =0.121212\cdots$$

Staying in base 3, we get

\begin{array}{rcr} N &= &0.121212\cdots \\ 100N &= &12.121212 \cdots \\ -N &= &-0.121212\cdots \\ \hline 22N &= &12.000000\cdots \\ \end{array}

So $N = \dfrac{12}{22}$

In base $10$, this becomes $N = \dfrac 58$

Added 3/19/2021

You can also do this directly

\begin{align} N &= \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + … \\ 9N &= 3 + 2 + \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + … \\ 9N &= 5 + N \\ N &= \frac 58 \end{align}

Answer 7

One more option that apparently hasn't been mentioned yet: in your original question you mentioned being stuck on the term $$\sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$ This can be rewritten as $$\frac{1}{2}\sum_{n=1}^{ \infty } \left(-\frac{1}{3}\right)^n$$ which can also be evaluated as a geometric series.